A) Constant
B) Proportional to t
C) Proportional to v
D) Proportional to s
Correct Answer: A
Solution :
\[s=\frac{1}{2}vt\] Þ \[2s=vt\]Þ \[2\frac{ds}{dt}=v+t.\frac{dv}{dt}\] Þ \[2\frac{{{d}^{2}}s}{d{{t}^{2}}}=\frac{dv}{dt}+t.\frac{{{d}^{2}}v}{d{{t}^{2}}}+\frac{dv}{dt}\] But \[\frac{dv}{dt}\]= acceleration Þ \[2a=a+t.\frac{da}{dt}+a\] Þ\[\frac{da}{dt}=0\] or \[xy=4\times 4=16\] But for whole notation \[t=0\]is impossible so that \[\frac{da}{dt}=0\] i.e., a is constant.
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